Question

Tough one!
For what values of a and b is the following equation true?
limx->0 ((sin2x)/x^3)+a+(b/x^2)) = 0
@Calculus1

Answer 1

it would have to amount to limit of the sums maybe

lim (sin2x)/x^3
lim a
lim b/x^2

Answer 2

Exactly! Good, I was on the right track. Okay so:

I figured lim sin2x/x^3 = 2/0
lim a=a
lim b/x^2= 0/2

Answer 3

the limit of a = a ....

so when does the lim of b/x^2 + lim sin(2x)/x^3 = -a

Answer 4

Wait, is that a rhetorical question? Was I wrong? Is the limit of a not a?

Answer 5

the easiest thing to do would be to get a limit for sin(2x)/x^3 to calibrate this stuff with

Answer 6

the limit of a constant is the constant

Answer 7

the whole stuff has to equal 0, so the other stuff has to amount to -a

Answer 8

a - a = 0

Answer 9

So Do i just plug in the limits I found for the other two parts of the equation?

Answer 10

aka 2/0 and 0/2, respectively

Answer 11

h = sin(2x)/x^3

h' = 2cos(2x)/3x^2

h'' = -4sin(2x)/6x

h''' = -8cos(2x)/6

lim h''' = -8 cos(0)/6 = -4/3 right?

Answer 12

ugh .... thought lhopital would do it; but the wolf says that goes up to infinity

Answer 13

lol, "the wolf"

Answer 14

you sure you got it typed up right?

http://www.wolframalpha.com/input/?i=solve+for+a+and+b+lim_%28x+to+0%29+%28%28sin2x%29%2Fx%5E3%29%2Ba%2B%28b%2Fx%5E2%29%29+%3D+0

it says no solutions

Answer 15

yea

Answer 16

Look what I found though.... Lol yahoo answers: http://answers.yahoo.com/question/index?qid=20091110070715AAcJ0Vq

Answer 17

Looks like you were right with -4/3. How come the wolf is saying that is wrong though..

Answer 18

thats where my lhopital went bad, i kept at it till the bottom was clear ....

Answer 19

Oh, that isn't allowed?

Answer 20

apparently not, its not in the correct indeterminate form to reapply it

Answer 21

Any ideas or are you stumped?

Answer 22

the wolf says those solutions from yahoo are false

Answer 23

the answer is sometimes: does not exist ....

Answer 24

Lets see if agreene can enlighten us?

Answer 25

since the limit fails to exist on the sin part, there is nothing you can do

Answer 26

Since:
\[\lim_{x \rightarrow 0} \frac{\sin (2x)}{x^3}= \infty\]
I should think that in order for this to approach 0...
a and/or b must be approaching negative infinity at the same rate(s).

Answer 27

a cant change since its constant

Answer 28

unless its 0

Answer 29

but then infinity - infinity is not defined to be zero

Answer 30

indeed, thats why I said it has to approach -infty at the same rate (to perpetually cancel the movement towards infinity)

I don't know of any numbers you can plug in to make this limit 0...

Answer 31

well, a=0 and b=-1 seems to work

Answer 32

http://www.wolframalpha.com/input/?i=lim_%28x+to+0%29+%28%28sinx%29%2Fx%5E3%29%2B0%2B+-1%2Fx%5E2+%3D+0

Answer 33

Interesting that... since:
http://www.wolframalpha.com/input/?i=limx-%3E0+%28%28sin2x%29%2Fx%5E3%29%2B0%2B%281%2Fx%5E2%29%29+

Answer 34

Oh i'm dumb sorry, I used pos 1.

Answer 35

O_o
didnt change anything with -1
http://www.wolframalpha.com/input/?i=limx-%3E0+%28%28sin2x%29%2Fx%5E3%29%2B0-%281%2Fx%5E2%29%29+

Answer 36

i spose inf-inf doesnt equal 0, unless when it does ....

Answer 37

wrap your equation

Answer 38

at the moment its taking lim sin and adding it to the rest

Answer 39

wrapping doesnt make a difference... I dont know why it's calling it 0 for you and infty for me O_o
http://www.wolframalpha.com/input/?i=limx-%3E0+%28%28%28sin2x%29%2Fx%5E3%29%2B0-%281%2Fx%5E2%29%29

Answer 40

lol .... i asked it it = 0 and it says true; then shows steps. the wolf has gone schitzoid i spose

Answer 41

NOOOOOOOOOOOOOOOOOOOO

Answer 42

I see why, you used sin(x)

It's sin(2x)...
sin(x) makes it go to 0
sin(2x) makes it go to infty.

Answer 43

i just noticed that too ....

Answer 44

fine. a=0, b=-2 lol

Answer 45

http://www.wolframalpha.com/input/?i=lim_%28x+to+0%29+%28%28sin2x%29%2Fx%5E3%29%2B0-2%2Fx%5E2%29+%3D+0

Answer 46

when i leave off the =0 , it says the limit is -4/3 lol

Answer 47

To answer the question, I think:
\[\forall a \lim f(x) \neq 0\]
and
\[\forall a \lim f(x) \neq 0\]

Even though I dont have any proof, lol

Answer 48

2nd one is supposed to be forall b

Answer 49

what does that V symbol mean? Is that an angle symbol?

Answer 50

Its an upside down A... it means For all (variable) there should be a :: after, but I forgot those.

Math shorthand is useful... instead of writing:
For all a such that limit of the function doesnt equal 0
you can just write:
\[\forall a :: \lim f(x) \neq 0\]

Answer 51

Wait so that means, "for all a such that the limit of the function doesn't equal 0", okay

Answer 52

Some people prefer | instead of :: for such that, I was taught with :: but most people use | nowdays.. meh

Answer 53

They come in handy:
http://en.wikipedia.org/wiki/Table_of_mathematical_symbols

Answer 54

It basically says, there is no A or B that exist by that will make it equal 0.

Answer 55

if a=0 and b=-2 we can lhopital this into submission i believe to get
\[\frac{sin(2x)}{x^3}-\frac{2}{x^2}\]
\[\frac{x^2sin(2x)-2x^3}{x^5},\frac{0}{0}\]
\[\frac{2xsin(2x)+2x^2 cos(2x)-6x^2}{5x^4},\frac{0}{0}\]
\[\frac{2sin(2x)+8xcos(2x)-4x^2 sin(2x)-12x}{20x^3},\frac{0}{0}\]
\[\frac{-6sin^2(x)-6xsin(2x)-2x^2cos(2x)}{15x^2},\frac{0}{0}\]
\[\frac{-6sin(2x)-8xcos(2x)+2x^2sin(2x)}{15x},\frac{0}{0}\]
\[\frac{-20cos(2x)+20xsin(2x)+4x^2cos(2x)}{15},x=0,-\frac{20}{15}=-\frac{4}{3}\]

Answer 56

which aint 0, but of a = 4/3 then it might just work. I cant trust the wolf with this since it has no idea what to do with it :)

Answer 57

it shows its good on the graph

Answer 58

the limit from the left matches the limit from the right ....

Answer 59

but then, thats only 1 value of a and b, i gots no idea what it would be like for "all" of them